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Effect of Doping on Depletion Region Width

  1. Aug 1, 2017 #1
    In an unbiased p-n junction, how does increase in doping affect the width of Depletion Region?

    I read somewhere that width decreases as increase in carrier concentration leads to more recombination of majority carriers with oppositely charged ions of their depletion regions leading to a decrease in depletion width.

    But in the case of high doping , there will be a greater concentration gradient of charge carriers which should lead to a greater amount of diffusion current. A higher diffusion current implies more no. of ions on either side of the junction which in turn should increase the depletion width.

    How can we say then that depletion width decreases with increase in doping?
     
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  3. Aug 1, 2017 #2

    scottdave

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    It has been several years, since I took that class or worked with this. So I did a search, to see what I could find. I found the following link, which talks about both the concentration gradient and the field which effect the size of the depletion region. There will be an equilibrium size where the effect of the field balances out the effect of the concentration gradient. After reading this, some of it started to come back to me. You should do more research on your own, but I believe this link explains it fairly well. https://physics.stackexchange.com/questions/88573/change-in-the-width-of-depletion-layer-with-doping
     
  4. Aug 1, 2017 #3

    analogdesign

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    We can say depletion width decreases with increase in doping because it does but only relative to the doping on the other side of the pn junction. If each side of the junction has equal numbers of majority carriers then the same depletion region width is needed to uncover the same amount of space charge. If one side is highly doped compared to the other side, then on the lightly spaced side more space charge is needed to balance the diffusion current. Therefore, it's depletion width is longer. Make sense?
     
  5. Aug 2, 2017 #4
    what itsaid seems reasonable but why didn't it take into account the fact that on increasing the doping , concentration gradient will also increase which will further increase the diffusion current or we can say that increased gradient will increase the potential of carriers to diffuse towards lower concentration regions?
     
  6. Aug 2, 2017 #5
    That seems to make sense but what i asked for is something else.

    Say for a Zener Diode , we require High Electric field in depletion region. And it can achieved by narrowing the width of depletion region.
    For this, they have mentioned that high doping on both sides of the junction can narrow the depletion width than what we obtain in case of moderate doping.
    How does it happen?

    P.S: The book i am referring to is "Fundamentals of Microelectronics" by B.Razavi.
     
  7. Aug 2, 2017 #6

    analogdesign

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    Again, everything is relative. Electric field is the spatial derivative of voltage, so higher doping means the voltage is dumped quicker, and that is just because more space charge per unit volume is uncovered. It is as simple as that. You have to drop a fixed voltage, to do it in a smaller region you need more charge, so you need more doping. It isn't deep physics going on here.
     
  8. Aug 2, 2017 #7
    Absolutely correct! No doubt in this, neither i had any ever over this theory. What i wanted to convey is that why don't you take into account the increased diffusion current due to high doping?
    Lets have 2 cases of unbiased p-n junctions:-
    1 : Moderate doping , concentration gradient=c1, Diffusion current=i1 , Depletion width=w1 , Electric field required to stop diffusion= e1
    2: High doping , concentration gradient=c2 ,Diffusion current=i2 , Depletion width=w2 , Electric field required to stop diffusion = e2

    Now, c2 > c1 => i2 > i1 => e2 > e1 (since diffusion current is more, required electric field should also be more)
    if required electric field was same in both cases , say e, no doubt w2 will be lesser than w1.
    but here, since e2 and e1 are not same, how can we comment on w2 < w1 ?

    p.s: Thanks for your time :)
     
  9. Aug 2, 2017 #8

    analogdesign

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    Yes, e2 > e1, therefore dV2/dx > dV1/dx therefore w2 < w1 since charge is balanced.
     
  10. Aug 2, 2017 #9
    Yes dV2/dx > dV1/dx , but again ΔV2 is not equal to ΔV1. ( instead ΔV2 > ΔV1)
    Potential gradient is more in case 2 , but potential drop is also more in case 2 due to increased doping.
     
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